There are three pigeon holes marked M, P, C. The number of ways in which we can put 12 letters so that 6 of them are in M, 4 are in P and 2 are in C is

There are three pigeon holes marked $M, P, C .$ The number of ways in which we can put 12 letters so that 6 of them are in $M,$ 4 are in $P$ and 2 are in $C$ is

A
2520
B
13860
C
12530
D
25220

Solution:

Number of ways is $^{12} C_{6}$ for $M,^{6} C_{4}$ for $P$ and $^{2} C_{2}$ for $C .$

Thus, the required number of ways

$= (^{12} C_{6}) (^{6} C_{4}) (^{2} C_{2})$

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